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A 320 g bird flying along at 7.0 m/s sees a 12 g insect heading straight toward it with a speed of 30 m/s (as measured by an observer on the ground, not by the bird). The bird opens its mouth wide and enjoys a nice lunch. What is the bird's speed immediately after swallowing?

Answer :

cjmejiab

To solve the problem it is necessary to apply the conservation equations of the moment.

The conservation equation of the moment is defined as,

[tex]m_1v_1+m_2v_2 = (m_1+m_2)v_f[/tex]

Where,

[tex]m_{1} =[/tex] Mass of Bird

[tex]m_2 =[/tex] Mass of Insect

[tex]v_{1} =[/tex]Velocity Bird before swallowing

[tex]v_2 =[/tex]Velocity of insect

[tex]v_f =[/tex] final velocity (both)

Replacing with our values we have,

[tex]m_1v_1+m_2v_2 = (m_1+m_2)v_3[/tex]

[tex](0.32)(7)+(0.012)(30)= (0.32+0.012)v_f[/tex]

[tex]2.6 = 0.332v_f[/tex]

[tex]v_f = 7.83m/s[/tex]

Therefore the bird's speed immediately after swallowing is 7.83m/s

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